An `AC` source is rated at `220V, 50 Hz.` The time taken for voltage to change from its peak value to zero is

Options

A. 50 sec
B. `0.02` sec
C. 5 sec
D. `0.005` sec

Answer

To solve the problem of finding the time taken for the voltage of an AC source to change from its peak value to zero, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the AC Voltage Waveform**:
The AC voltage can be represented as a sinusoidal wave. The peak voltage (maximum voltage) is denoted as \( V_m \) and occurs at \( t = 0 \).

2. **Identify the Time Period (T)**:
The time period \( T \) of the AC waveform is the time taken to complete one full cycle. The frequency \( f \) is given as 50 Hz. The relationship between time period and frequency is given by:
\[ T = \frac{1}{f} \]
Substituting the given frequency:
\[ T = \frac{1}{50} = 0.02 \text{ seconds} \]

3. **Determine the Time to Go from Peak to Zero**:
The time taken for the voltage to change from its peak value to zero is one-fourth of the time period \( T \). This is because the voltage waveform reaches zero at \( T/4 \) after reaching its peak. Therefore:
\[ \text{Time from peak to zero} = \frac{T}{4} \]

4. **Calculate the Time from Peak to Zero**:
Now substituting the value of \( T \):
\[ \text{Time from peak to zero} = \frac{0.02}{4} = 0.005 \text{ seconds} \]

5. **Final Answer**:
Thus, the time taken for the voltage to change from its peak value to zero is:
\[ 0.005 \text{ seconds} \text{ or } 5 \times 10^{-3} \text{ seconds} \]

### Summary:
The time taken for the voltage to change from its peak value to zero is \( 0.005 \) seconds.